Splitting field of $X^3-5$ and over $\Bbb{F}_p$, $p=7,11,13$

Question

Please help me to find the splitting field of $X^3-5$ over $\Bbb{F}_p$, $p=7,11,13$.

Thanks,

Answer 1

If a third degree polynomial hasn’t root on his field of definition it is irreducible.

Seeing $f(x)=x^3-5\in\mathbb{F}_7[x]$ you can see by counting that $\nexists y\in\mathbb{F}_7$ so that $y^3\equiv 5 \bmod7$ so the polinomial is irreducible and his split field is $\mathbb{F}_{7^3}$.

Instead in $\mathbb{F}_{11}$ $3$ is a root of $f(x)$, so you have $x^3-5=x^3-3^3=(x-3)(x^2+3x+9)$ where the quadratic has discriminant $\Delta =-27=6$ then it is reducible iff $6$ is a perfect square in $\mathbb{F}_{11}$ and by counting you see it is no possible. So the split field of $f(x)\in \mathbb{F}_{11}[x]$ is $\mathbb{F}_{11^2}$

Seeing $f(x)\in \mathbb{F}_{13}[x]$ we have that $2^3=\equiv -5\, \bmod13\iff (-2)^3\equiv 11^3\equiv 5\, \bmod13$. So we have that $11$ is root of $f(x)$ and $x^3-5=(x-11)(x^2+11x+121)=(x-11)(x^2+11x+4)$ where the quadratic has discriminant $\Delta =121-16=105=1$ so it is reducible and $f(x)$ is completely splittable on $\mathbb{F}_{13}$.

Answer 2

The polynomial $x^3-5$ doesn't have a root in $\Bbb F_7$, so is irreducible. Hence the splitting field is $\Bbb F_{7^3}$.

Over $\Bbb F_{13}$, we have a root ($11\cong -2$). Thus, using @Jack D'Aurizio's comments, the polynomial splits.